Question #704fa

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Answer 1 · 2017-10-12T23:52:28

$k=2$

For a quadratic equation to have equal roots (also known as a double root, or having a multiplicity of 2), the discriminant must equal 0.

So $b^2-4ac=0$

In this problem $a=(k-3), b=2(k-3), and c=-1$

Thus:

$(2(k-3))^2-4(k-3)(-1)=0$

$4(k^2-6k+9)+4k-12=0$

$4k^2-24k+36+4k-12=0$

$4k^2-20k+24=0$

$4(k^2-5k+6)=0$

$4(k-2)(k-3)=0$

So $k=2 or 3$

However $k=3$ is an extraneous solution because if you plug in $3$ into the original equation you end up getting $-1=0$ which is false.

So $k=2$ is the only solution.